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$$\int_{-\infty}^\infty \frac{dx}{x^3+1}$$

Online integral calculator says this definite integral is $\pi/3$ but my textbook says it diverges. Integrand is undefined at $x=-1$ and I don't know if we can use Cauchy principal value.

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  • $\begingroup$ We can use the principal value (the pole at $-1$ is simple). $\endgroup$ – Daniel Fischer May 25 '17 at 12:45
  • $\begingroup$ Ofcourse, you're talking about the residue theorem, but this question is in textbook of advanced calculus class, in the chapter convergence of integrals, so there must be another way to decide. $\endgroup$ – offret May 25 '17 at 12:46
  • $\begingroup$ The residue theorem is very convenient to evaluate the integral, but to see that a) the integral doesn't exist as an improper Riemann integral or a Lebesgue integral, and b) the Cauchy principal value of the integral exists is independent of the residue theorem, and accessible to students with a modicum of integration theory. Obvious estimates show that the integrals over $(-\infty,-1-\varepsilon]$ and over $[-1+\varepsilon,+\infty)$ exist for every $\varepsilon > 0$. $\endgroup$ – Daniel Fischer May 25 '17 at 13:00
  • $\begingroup$ Then writing $x^3+1 = (x+1)(x^2-x+1)$ shows that the integrand is nearly "odd about $-1$", so one isn't surprised that the infinite parts cancel. More formally, one shows that the limits as $\varepsilon \downarrow 0$ of $$\int_{-2}^{-1-\varepsilon} \frac{1}{x^3+1} - \frac{1}{3(x+1)}\,dx\quad\text{and}\quad \int_{-1+\varepsilon}^0 \frac{1}{x^3+1} - \frac{1}{3(x+1)}\,dx$$ exist, because $$\frac{1}{x^3+1} - \frac{1}{3(x+1)}=\frac{3 - (x^2-x+1)}{3(x+1)(x^2-x+1)} = \frac{(x+1)(2-x)}{3(x+1)(x^2-x+1)} = \frac{2-x}{3(x^2-x+1)}$$ is continuous (after removing the removable singularity) on $\mathbb{R}$. $\endgroup$ – Daniel Fischer May 25 '17 at 13:00

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